There is only one real value of a' for which the quadratic equation ax2+(a+3)x+a−3=0 has two positive integral solutions. The solutions is

Sum of the roots =−(a+3)a=I+a=−3I++1-----iProduct of the roots =αβ=a−3a=I++2----iiand D=(a+3)2−4a(a+3) =9(I++1)2{(I+−2)2−12}D must be perfect square, then I+=6From Eq. (i),Product of the roots =I++2=6+2=8